Question

A random sample of 30 households was selected as part of a study on electricity usage, and the number of kilowatt-hours (kWh) was recorded for each household in the sample for the last quarter of 2010. The average usage was found to be 450 kWh. In a very large study in the last quarter of the previous year, it was found that the standard deviation of the usage was 81 kWh. Assuming that the standard deviation is unchanged and that the usage is normally distributed, an expression for calculating a 99% confidence interval for the mean usage in the last quarter of 2010 is:

Answer #1

Solution :

Given that,

= 450

s = 81

n = 30

Degrees of freedom = df = n - 1 = 30 - 1 = 29

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2,df = t0.005,29 = 2.756

Margin of error = E = t/2,df * (s /n)

= 2.756 * ( 81/ 30)

= 40.757

The 99% confidence interval estimate of the population mean is,

- E < < + E

450 - 40.757 < < 450 + 40.757

409 < < 491

(409,491)

**99% confidence interval for the mean usage in the last
quarter of 2010 is 409kwh to 491kwh.**

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